options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 32.8 , b ) 36 , c ) 32.1 , d ) 32.2 , e ) 32.9 | b | add(divide(circumface(7), const_2), multiply(7, const_2)) | the radius of a semi circle is 7 cm then its perimeter is ? | "radius = 7 cm then diameter = 14 cm 1 / 2 * 22 / 7 * 14 + 14 = 36 answer : b" | a = circumface / (
b = a + 2
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a ) rs 108000 , b ) rs 144000 , c ) rs 132000 , d ) rs 152400 , e ) none of these | d | add(multiply(multiply(add(multiply(const_2, 400), multiply(40, subtract(const_10, const_1))), const_12), add(const_2, const_3)), multiply(multiply(add(multiply(const_2, 500), multiply(subtract(20, const_1), 20)), 6), const_10)) | two persons raj and ramu started working for a company in similar jobs on january 1 , 1991 . raj ' s initial monthly salary was rs 400 , which increases by rs 40 after every year . ramu ' s initial monthly salary was rs 500 which increases by rs 20 after every 6 months . if these arrangements continue till december 31 , 200 . find the total salary they received during that period . | explanation : raj ' s salary as on 1 jan 1991 is rs 400 per month . his increment in his month salary is rs 40 per annum his total salary from 1 jan 1991 to 31 st dec 2000 i . e . in ten years = > 12 [ 2 ( 400 ) + ( 10 − 1 ) 40 ] × 10 / 2 . = > rs 69600 . ramu ' s salary as on jan 1 st 1991 is rs 550 and his half yearly increment in his month salary is rs 20 . his total salary from 1 jan 1991 to dec 31 , 2000 = > 6 [ 2 ( 500 ) + ( 20 − 1 ) 20 ] × 20 / 2 . = > rs 82000 . total salary of raj and ramu in the ten year period : = > rs . 69600 + rs . 82800 . = > rs 1 , 52400 . answer : d | a = 2 * 400
b = 10 - 1
c = 40 * b
d = a + c
e = d * 12
f = 2 + 3
g = e * f
h = 2 * 500
i = 20 - 1
j = i * 20
k = h + j
l = k * 6
m = l * 10
n = g + m
|
a ) 25 % , b ) 20 % , c ) 22 % , d ) 21 % , e ) 24 % | e | divide(multiply(divide(multiply(80, 60), const_100), 50), const_100) | at a certain college , 80 percent of the total number of students are freshmen . if 60 percent of the fresh - men are enrolled in the school of liberal arts and , of these , 50 percent are psychology majors , what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? | let ' s say there is a total of 100 students at this college . 80 percent of the total number of students are freshmen . # of freshmen = 80 % of 100 = 80 60 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = 60 % of 80 = 48 . . . and , of these , 50 percent are psychology majors . . . number of liberal arts freshmen who are psychology majors = 50 % of 48 = 24 what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? 24 / 100 = 24 % answer : e | a = 80 * 60
b = a / 100
c = b * 50
d = c / 100
|
a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2490 , 4150 , e ) 2505 , 4175 | e | multiply(divide(7.5, 12.5), divide(1670, subtract(const_1, divide(7.5, 12.5)))) | difference of two numbers is 1670 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | "let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1670 5 * y / 3 – y = 1670 2 * y / 3 = 1670 y = [ ( 1670 * 3 ) / 2 ] = 2505 . one number = 2505 , second number = 5 * y / 3 = 4175 . answer e ." | a = 7 / 5
b = 7 / 5
c = 1 - b
d = 1670 / c
e = a * d
|
a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | d | multiply(divide(4, 8), multiply(const_4, 1)) | the ratio of flour to water to sugar in a recipe is 11 : 8 : 1 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 4 cups of water , how much sugar is required ? | "the ratio of flour to water is 22 : 8 = 11 : 4 . the ratio of flour to sugar is 5.5 : 1 = 11 : 2 . the new ratio of flour to water to sugar is 11 : 4 : 2 if we need 4 cups of water , then we need 2 cups of sugar . the answer is d ." | a = 4 / 8
b = 4 * 1
c = a * b
|
a ) $ 412 , b ) $ 526 , c ) $ 585 , d ) $ 605 , e ) $ 662 | e | add(divide(3620, 10), multiply(50, 6)) | craig sells major appliances . for each appliance he sells , craig receives a commission of $ 50 plus 10 percent of the selling price . during one particular week craig sold 6 appliances for selling prices totaling $ 3620 . what was the total of craig ' s commissions for that week ? | craig receives a commission of $ 50 on each appliance plus a 10 % commission on total sales , thus his commission was 6 * 50 + 0.1 * 3620 = 662 . answer : e . | a = 3620 / 10
b = 50 * 6
c = a + b
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(9, const_3) | if 8 ^ x = 2 ^ 9 , what is x ? | 2 ^ 3 x = 2 ^ 9 3 x = 9 , therefore x = 3 answer b | a = 9 / 3
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a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 3 / 23 | e | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 3), 21), const_4.0), const_2), 3), const_3)) | a certain list consists of 21 different numbers . if n is in the list and n is 3 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction of the sum of the 21 numbers in the list ? | "series : a 1 , a 2 . . . . a 20 , n sum of a 1 + a 2 + . . . + a 20 = 20 * x ( x = average ) so , n = 3 * x hence , a 1 + a 2 + . . + a 20 + n = 23 x so , the fraction asked = 3 x / 23 x = 3 / 23 answer e" | a = 1 * 1
b = 20 / 3
c = b + 21
d = c / 4
e = d * 2
f = e - 3
g = f - 3
h = a / g
|
a ) 600 , b ) 700 , c ) 800 , d ) 900 , e ) 848 | e | multiply(add(100, 6), 8) | i chose a number and divide it by 8 . then i subtracted 100 from the result and got 6 . what was the number i chose ? | "solution : let x be the number i chose , then x / 8 â ˆ ’ 100 = 6 x / 8 = 106 x = 848 answer e" | a = 100 + 6
b = a * 8
|
a ) 128 , b ) 256 , c ) 512 , d ) 1024 , e ) 2048 | b | subtract(power(2, add(7, const_1)), const_1) | the population of a bacteria colony doubles every day . if it was started 7 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ? | 2 ^ 7 ( 2 ) = 2 ^ 8 = 256 the answer is b . | a = 7 + 1
b = 2 ** a
c = b - 1
|
a ) 180 times , b ) 381 times , c ) 155 times , d ) 392 times , e ) 150 times | c | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 32) | light glows for every 32 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 32 ) = 15582 times . answer : c" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 32
|
['a ) 3 √ 2', 'b ) 5', 'c ) 5 √ 2', 'd ) √ ( 67 )', 'e ) √ ( 73 )'] | b | sqrt(add(power(divide(8, const_2), const_2), power(3, const_2))) | if the height of a right cone c is 3 and the diameter of the base of c is 8 , what is the distance from the apex of c to any point on the circumference of the base of c ? | height of cone = 3 radius of base = 4 so the distance from the apex of c to any point on the circumference of the base of c will be a right angled triangle hence the distance is sqrt ( 3 ^ 2 + 4 ^ 2 ) = 5 answer : b | a = 8 / 2
b = a ** 2
c = 3 ** 2
d = b + c
e = math.sqrt(d)
|
a ) 11 hours , b ) 12 hours , c ) 13 hours , d ) 14 hours , e ) 18 hours | c | divide(multiply(26, 10), subtract(70, 50)) | a can copy 50 papers in 10 hrs , while a & b can copy 70 papers in 10 hrs . how many hours are required for b to copy 26 papers ? | a can copy 50 papers in 10 hrs , while a & b can copy 70 papers in 10 hrs . it means b can copy 20 papers in 10 hrs . then time taken by b to copy 26 papers = 26 * 10 / 20 = 13 hours answer : c | a = 26 * 10
b = 70 - 50
c = a / b
|
a ) 330 , b ) 440 , c ) 240 , d ) 550 , e ) 230 | d | divide(multiply(45, 816), 23) | 45 % of 816 - 23 % of ? = 240.7 | "d 550 ( 816 * 45 ) / 100 - ( ? * 23 ) / 100 = 240.7 = > 367.2 - ( ? * 23 ) / 100 = 240.7 = > ( ? * 23 ) / 100 = 367.2 - 240.7 = > ? = ( 126.5 * 100 ) / 23 = 550" | a = 45 * 816
b = a / 23
|
a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these | e | divide(sqrt(100), sqrt(121)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 121 hours and 100 hours respectively . the ratio of their speeds is : | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 100 : â ˆ š 121 = 10 : 11 . answer e" | a = math.sqrt(100)
b = math.sqrt(121)
c = a / b
|
a ) 6999.2 , b ) 6876.1 , c ) 6654 , d ) 7000 , e ) none of these | b | add(subtract(6650, divide(multiply(6650, 6), const_100)), divide(multiply(subtract(6650, divide(multiply(6650, 6), const_100)), const_10), const_100)) | shanmukham buys good worth rs . 6650 . he gets a rebate of 6 % on it . after getting the rebate , he pays sales tax @ 10 % . find the amount he will have to pay for the goods | explanation : rebate = 6 % of rs . 6650 = rs . 6 / 100 x 6650 = rs . 399 . sales tax = 10 % of rs . ( 6650 399 ) = rs . 10 / 100 x 6251 = rs . 625.10 final amount = rs . ( 6251 + 625.10 ) = rs . 6876.10 answer : b | a = 6650 * 6
b = a / 100
c = 6650 - b
d = 6650 * 6
e = d / 100
f = 6650 - e
g = f * 10
h = g / 100
i = c + h
|
a ) 71 , b ) 910 , c ) 1001 , d ) 1911 , e ) none | a | subtract(781, 710) | the maximum numbers of students among them 781 pens and 710 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is | "olution required number of students . = h . c . f of 781 and 710 . â € ¹ = â € º 71 . answer a" | a = 781 - 710
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a ) 8.5 units , b ) 6 units , c ) 3 units , d ) 5 units , e ) 2 units | e | divide(triangle_area_three_edges(8, 15, 27), divide(triangle_perimeter(8, 15, 27), const_2)) | what is the measure of the radius of the circle inscribed in a triangle whose sides measure 8 , 15 and 27 units ? | "sides are 8 , 15 and 27 . . . thus it is right angle triangle since 27 ^ 2 = 8 ^ 2 + 15 ^ 2 therefore , area = 1 / 2 * 15 * 8 = 60 we have to find in - radius therefore , area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = 27 + 15 + 8 / 2 = 30 thus , 60 = 30 * r and hence r = in - radius = 2 option e" | a = triangle_area_three_edges / (
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a ) 58 , b ) 59 , c ) 62 , d ) 66 , e ) 74 | d | add(add(multiply(divide(60, 3), const_2), const_3), add(divide(60, 3), const_3)) | the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 3 years hence ? | explanation : a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 3 years , their ages will be 43 and 23 . sum of their ages = 43 + 23 = 66 . d ) | a = 60 / 3
b = a * 2
c = b + 3
d = 60 / 3
e = d + 3
f = c + e
|
a ) 10.9 sec , b ) 11.6 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 20.8 sec | b | divide(add(120, 190), multiply(add(60, 40), const_0_2778)) | two trains 120 m and 190 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 120 + 190 = 310 m . required time = 310 * 9 / 250 = 11.6 = 11.6 sec . answer : b" | a = 120 + 190
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 4000 , b ) 1000 , c ) 5000 , d ) 3000 , e ) 2000 | e | divide(26000, add(multiply(const_3, const_4), const_1)) | krishan and nandan jointly started a business . krishan invested 4 times as nandan did and invested his money for trible time as compared to nandan . if the gain is proportional to the money invested and the time for which the money is invested and the total gain was rs . 26000 , find the nandan ' s earning . | 4 : 1 3 : 1 - - - - - - 12 : 1 13 - - - - - rs . 26000 1 - - - - - ? = > rs . 26000 / 13 = rs . 2000 answer : e | a = 3 * 4
b = a + 1
c = 26000 / b
|
a ) 0 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | d | divide(1, 37) | what is the 21 th digit to the right of the decimal point in the decimal expansion of 1 / 37 ? | "1 / 37 = 0.027027 . . . so , we have a repeating cycle of 027 . every third digit ( 3 rd , 6 th , 9 th , . . . ) to the right of the decimal point is 7 , thus 21 st digit to the right of the decimal point is also 7 . answer : d ." | a = 1 / 37
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a ) 2 / 35 , b ) 2 / 3 , c ) 7 / 35 , d ) 5 / 7 , e ) 7 / 5 | a | multiply(divide(2, 7), divide(1, 5)) | two brothers ram and ravi appeared for an exam . the probability of selection of ram is 2 / 7 and that of ravi is 1 / 5 . find the probability that both of them are selected . | "let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 2 / 7 p ( b ) = 1 / 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 2 / 7 x 1 / 5 = 2 / 35 answer : a" | a = 2 / 7
b = 1 / 5
c = a * b
|
a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | d | divide(subtract(divide(42, const_2), sqrt(subtract(power(divide(42, const_2), const_2), multiply(const_4, 104)))), const_2) | if a rectangular billboard has an area of 104 square feet and a perimeter of 42 feet , what is the length of each of the longer sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 104 and a perimeter of 42 . we ' re asked for the length of one of the longer sides of the rectangle . since the answers are all integers , and the area is 104 , the shorter side will almost certainly be less than 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . answer b ( 7 ) does not divide evenly into 104 , so the correct answer is probably a or c . let ' s test answer c : 8 if . . . the shorter side = 8 . . . the area = 104 . . . . 104 / 8 = 13 = the longer side perimeter = 8 + 8 + 13 + 13 = 42 d" | a = 42 / 2
b = 42 / 2
c = b ** 2
d = 4 * 104
e = c - d
f = math.sqrt(e)
g = a - f
h = g / 2
|
a ) 9 % , b ) 12 % , c ) 10 % , d ) 36 % , e ) 40 % | a | multiply(divide(75, subtract(900, 75)), const_100) | a cricket bat is sold for $ 900 , making a profit of $ 75 . the profit percentage would be | "75 / ( 900 - 75 ) = 75 / 825 = 0.9 bit more than 9 % . answer : a" | a = 900 - 75
b = 75 / a
c = b * 100
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(4, add(2, 2)) | if 4 ^ ( 2 x + 2 ) = 16 ^ ( 3 x − 1 ) , what is the value of x ? | 2 ^ ( 4 x + 4 ) = 2 ^ ( 12 x - 4 ) 4 x + 4 = 12 x − 4 x = 1 answer b | a = 2 + 2
b = 4 / a
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a ) 11 , b ) 15 , c ) 17 , d ) 38 , e ) 56 | c | multiply(power(const_2, const_4.0), factorial(720)) | the product of three consecutive numbers is 720 . then the sum of the smallest two numbers is ? | "product of three numbers = 720 720 = 8 * 9 * 10 so , the three numbers are 8 , 9 and 10 . and sum of smallest of these two = 8 + 9 = 17 . answer : option c" | a = 2 ** 4
b = math.factorial(720)
c = a * b
|
a ) 39 , b ) 40 , c ) 41 , d ) 75 , e ) 43 | d | subtract(multiply(add(22, 25), const_2), 19) | you collect pens . suppose you start out with 25 . mike gives you another 22 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | "solution start with 25 pens . mike gives you 22 pens : 25 + 22 = 47 pens . cindy doubles the number of pens you have : 47 ã — 2 = 94 pens . sharon takes 19 pens from you : 94 - 19 = 75 pens . so you have 75 at the end . correct answer : d" | a = 22 + 25
b = a * 2
c = b - 19
|
a ) 4.37 % , b ) 5.8 % , c ) 6.8 % , d ) 8.75 % , e ) none | b | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 45,000 to 2 , 30,500 in a decade . the average percent increase of population per year is : | "explanation : increase in 10 years = ( 230500 - 145000 ) = 85500 . increase % = ( 85500 / 145000 x 100 ) % = 58 % . required average = ( 58 / 10 ) % = 5.8 % . answer : option b" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 2 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 2 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) 15 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | a | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 1) | in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 1 ounces of corn syrup , how many ounces of water does it contain ? | "standard : fl : corn s : water = 1 : 12 : 30 sport : fl : corn s : water = 3 : 12 : 180 this simplifies to 1 : 4 : 60 if the large bottle has a capacity of x ounces , then 4 x / 65 = 1 . so , x = 65 / 4 ounces . water = ( 60 / 65 ) * ( 65 / 4 ) = 15 ounces . ans a" | a = 1 / 12
b = a * 3
c = 1 / 30
d = c / 2
e = b / d
f = e * 1
|
a ) 15 , b ) 25 , c ) 30 , d ) 45 , e ) 60 | c | multiply(divide(divide(multiply(7, 8), subtract(11, 7)), 7), 15) | in a certain animal shelter , the ratio of the number of dogs to the number of cats is 15 to 7 . if 8 additional cats were to be taken in by the shelter , the ratio of the number of dogs to the number of cats would be 15 to 11 . how many dogs are in the shelter ? | "this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of dogs to the number of cats is 15 : 7 . we ' re then told that 8 more cats are added to this group and the ratio becomes 15 : 11 . we ' re asked for the number of dogs . algebraically , since the number of dogs is a multiple of 15 and the number of cats is a multiple of 7 , we can write this initial relationship as . . . 15 x / 7 x when we add the 18 cats and factor in the ' ending ratio ' , we have an equation . . . . 15 x / ( 7 x + 8 ) = 15 / 11 here we have 1 variable and 1 equation , so we can solve for x . . . . ( 15 x ) ( 11 ) = ( 7 x + 8 ) ( 15 ) ( x ) ( 11 ) = ( 7 x + 8 ) ( 1 ) 11 x = 7 x + 8 4 x = 8 x = 2 with this x , we can figure out the initial number of dogs and cats . . . initial dogs = 15 x = 15 ( 2 ) = 30 final answer : c" | a = 7 * 8
b = 11 - 7
c = a / b
d = c / 7
e = d * 15
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a ) 504 , b ) 532 , c ) 210 , d ) 180 , e ) 280 | d | add(divide(add(divide(523, const_4), divide(523, const_4)), const_2), multiply(add(const_4, const_1), const_10)) | the number 523 fbc is divisible by 7 , 89 . then what is the value of f * b * c | lcm of 7 , 8 and 9 is 504 , thus 523 fbc must be divisible by 504 . 523 fbc = 523000 + fbc 523000 divided by 504 gives a remainder of 352 . hence , 352 + fbc = k * 504 . k = 1 fbc = 152 - - > f * b * c = 10 k = 2 fbc = 656 - - > f * b * c = 180 as fbc is three digit number k can not be more than 2 . two answers ? well only one is listed in answer choices , so d . answer : d . | a = 523 / 4
b = 523 / 4
c = a + b
d = c / 2
e = 4 + 1
f = e * 10
g = d + f
|
a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 4.00 , e ) $ 5.00 | e | add(1.50, multiply(0.50, subtract(add(divide(subtract(18, 2), 2), const_1), 2))) | the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 1.50 + 0.50 ( x − 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 2 wheels on each of its other axles ? | "number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 2 = 8 total number of axles = 9 t = 1.50 + 0.50 ( 9 − 2 ) = 1.50 + . 5 * 7 = 1.5 + 1.5 = 5 $ answer e" | a = 18 - 2
b = a / 2
c = b + 1
d = c - 2
e = 0 * 50
f = 1 + 50
|
a ) 8 , b ) 2 , c ) 9 , d ) 5 , e ) 1 | a | subtract(subtract(multiply(5000, power(add(const_1, divide(4, const_100)), 2)), 5000), multiply(multiply(5000, divide(4, const_100)), 2)) | indu gave bindu rs . 5000 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "5000 = d ( 100 / 4 ) 2 d = 8 answer : a" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 5000 * c
e = d - 5000
f = 4 / 100
g = 5000 * f
h = g * 2
i = e - h
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a ) $ 508 , b ) $ 698 , c ) $ 398 , d ) $ 549 , e ) $ 675 | a | subtract(700, divide(multiply(subtract(764, 700), 3), 4)) | a sum of money at simple interest amounts to $ 700 in 3 years and to $ 764 in 4 years . the sum is : | "a $ 508 s . i . for 1 year = $ ( 764 - 700 ) = $ 64 . s . i . for 3 years = $ ( 64 x 3 ) = $ 192 . principal = $ ( 700 - 192 ) = $ 508 ." | a = 764 - 700
b = a * 3
c = b / 4
d = 700 - c
|
a ) 2 , b ) 8 , c ) 20 , d ) 25 , e ) 26 | c | divide(34, multiply(const_10, const_2)) | how many factors does 34 ^ 2 have ? | "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 5 * 4 = 20 answer c ." | a = 10 * 2
b = 34 / a
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a ) 5 , b ) 4 , c ) 7 , d ) 1 , e ) 6 | e | divide(subtract(100, 1), 15) | how many positive integers between 1 and 100 are there such that they are multiples of 15 ? | "multiples of 15 = 15 , 30,45 , - - - - - 90 number of multiples of 15 = > 90 - 15 / 15 + 1 = 6 answer is e" | a = 100 - 1
b = a / 15
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a ) 500 , b ) 650 , c ) 250 , d ) 111 , e ) 236 | a | multiply(divide(multiply(10, const_1000), const_60), 3) | a man walking at a rate of 10 km / hr crosses a bridge in 3 minutes . the length of the bridge is ? | "speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 5 minutes = 50 / 18 * 5 * 60 = 500 m answer is a" | a = 10 * 1000
b = a / const_60
c = b * 3
|
a ) 99 , b ) 18 , c ) 26 , d ) 15 , e ) 12 | d | divide(rectangle_area(15, 6), rectangle_area(3, 2)) | how many paying stones , each measuring 3 * 2 m are required to pave a rectangular court yard 15 m long and 6 m board ? | "15 * 6 = 3 * 2 * x = > x = 15 answer : d" | a = rectangle_area / (
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a ) 1 / 4 , b ) 4 / 5 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | e | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction w of the total distance did he cover while the sun was shining on him ? | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or 12 t . therefore , total distance covered while the sun was shining over him is w = 2 / 14 = 1 / 7 . answer : e | a = math.lcm(2, 3)
b = a / 2
c = b - 2
|
a ) 130 , b ) 100 , c ) 200 , d ) 150 , e ) 120 | d | multiply(10, 15) | what number has a 15 : 1 ratio to the number 10 ? | "15 : 1 = x : 10 x = 10 * 15 x = 150 answer : d" | a = 10 * 15
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a ) 16 , b ) 32 , c ) 64 , d ) 256 , e ) 128 | d | multiply(power(2, const_3.0), 2) | if a and b are positive integers and ( 2 ^ a ) ^ b = 2 ^ 7 , what is the value of 2 ^ a * 2 ^ b ? | "2 ^ ab = 2 ^ 7 therefore ab = 7 either a = 1 or 7 or b = 7 or 1 therefore 2 ^ a * 2 ^ b = 2 ^ ( a + b ) = 2 ^ 8 = 256 d" | a = 2 ** 3
b = a * 2
|
a ) 22 years , b ) 23 years , c ) 24 years , d ) 25 years , e ) 26 years | e | divide(subtract(30, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 30 years older than his son . in four years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 30 . = > 2 ( x + 4 ) = ( x + 30 + 4 ) = > 2 x + 8 = x + 34 = > x = 26 years answer : option e" | a = 2 * 2
b = a - 2
c = 30 - b
d = 2 - 1
e = c / d
|
a ) 300 , b ) 120 , c ) 150 , d ) 170 , e ) 270 | d | divide(multiply(20, 17), const_2) | the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 17 terms of the arithmetic progression ? | "n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 17 . . . we get 17 / 2 [ 2 a + 14 d ] = 17 [ a + 7 d ] = 17 * 10 = 170 . . . answer is d . . ." | a = 20 * 17
b = a / 2
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a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 2000 , e ) none of these | d | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | a man invests rs . 4,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | "explanation : interest on rs . 4000 at 5 % per annum = ( 4000 × 5 × 1 ) / 100 = rs . 200 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 4000 + x ) × 6 × 1 / 100 . = > 200 + 2 x / 25 = ( 4000 + x ) × 6 × 1 / 100 . = > 20000 + 8 x = ( 4000 + x ) × 6 . = > 20000 + 8 x = 24000 + 6 x . = > 2 x = 4000 . = > x = 2000 . answer : d" | a = 8 * 1000
b = 6 - 5
c = 8 - 6
d = b / c
e = a * d
|
a ) $ 400 , b ) $ 420 , c ) $ 440 , d ) $ 460 , e ) $ 480 | c | divide(add(add(add(add(400, 250), 650), 400), 500), 5) | a cab driver 5 days income was $ 400 , $ 250 , $ 650 , $ 400 , $ 500 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 400 + 250 + 650 + 400 + 500 ) / 5 = 440 answer is c" | a = 400 + 250
b = a + 650
c = b + 400
d = c + 500
e = d / 5
|
a ) 21.2 seconds , b ) 25.2 seconds , c ) 29.2 seconds , d ) 35.2 seconds , e ) 11.2 seconds | b | divide(add(120, 160), divide(40, const_3_6)) | calculate the time it will take for a train that is 120 meter long to pass a bridge of 160 meter length , if the speed of the train is 40 km / hour ? | speed = 40 km / hr = 40 * ( 5 / 18 ) m / sec = 11.1111 m / sec total distance = 120 + 160 = 280 meter time = distance / speed = 280 * ( 1 / 11.1111 ) = 25.2 seconds answer : b | a = 120 + 160
b = 40 / const_3_6
c = a / b
|
a ) 1240 , b ) 1120 , c ) 1190 , d ) 1290 , e ) none of these | d | add(540, divide(multiply(540, const_100), multiply(12, 6))) | the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 540 . what is the banker ' s discount ? | "explanation : td = ( bg × 100 ) / tr = ( 540 × 100 ) / ( 6 × 12 ) = ( 90 × 100 ) / 12 = ( 15 × 100 ) 2 / = rs . 750 bg = bd – td = > 540 = bd - 750 = > bd = 540 + 750 = 1290 answer : option d" | a = 540 * 100
b = 12 * 6
c = a / b
d = 540 + c
|
a ) 40 , b ) 50 , c ) 54 , d ) 60 , e ) none of these | d | divide(multiply(20, 30), subtract(30, 20)) | a and b can together finish a work in 30 days . they worked together for 20 days and then b left . after another 20 days , a finished the remaining work . in how many days a alone can finish the job ? | "( a + b ) ' s 20 days work = ( 1 / 30 * 20 ) = 2 / 3 . remaining work = ( 1 - 2 / 3 ) = 1 / 3 . now , 1 / 3 work is done by a in 20 days . whole work will be done by a in ( 20 * 3 ) = 60 days . correct option : d" | a = 20 * 30
b = 30 - 20
c = a / b
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | subtract(floor(divide(580, lcm(lcm(const_4, add(const_4, const_1)), 6))), floor(divide(190, lcm(lcm(const_4, add(const_4, const_1)), 6)))) | how many numbers between 190 and 580 are divisible by 4,5 and 6 ? | every such number must be divisible by l . c . m of 4 , 5,6 i . e , 60 . such numbers are 240,300 , 360,420 , 480,540 . clearly , there are 6 such numbers answer : a | a = 4 + 1
b = math.lcm(4, a)
c = math.lcm(b, 6)
d = 580 / c
e = math.floor(d)
f = 4 + 1
g = math.lcm(4, f)
h = math.lcm(g, 6)
i = 190 / h
j = math.floor(i)
k = e - j
|
a ) 389 m , b ) 350 m , c ) 187.5 m , d ) 299 m , e ) 219.5 m | c | subtract(multiply(speed(300, 24), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 24 sec . what is the length of the platform ? | "speed = 300 / 24 = 25 / 2 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 25 / 2 = > x = 187.5 m . answer : c" | a = speed * (
b = a - 39
|
a ) 60 % , b ) 50 % , c ) 75 % , d ) 45 % , e ) 30 % | a | multiply(divide(3, 5), const_100) | what percent is 3 % of 5 % ? | "required percentage = 3 % / 5 % * 100 = 3 / 5 * 100 = 60 % answer is a" | a = 3 / 5
b = a * 100
|
a ) 560 , b ) 616 , c ) 672 , d ) 728 , e ) 784 | b | divide(59.32, subtract(2, floor(2))) | when positive integer x is divided by positive integer y , the result is 59.32 . what is the sum t of all possible 2 - digit remainders for x / y ? | "ans b 616 . . . remainders = . 32 = 32 / 100 = 8 / 25 = 16 / 50 and so on . . so two digit remainders are 16 + 24 + 32 + . . . . + 96 . . t = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616 . b" | a = math.floor(2)
b = 2 - a
c = 59 / 32
|
['a ) 7', 'b ) 8', 'c ) 9', 'd ) 10', 'e ) 11'] | d | subtract(40, multiply(divide(45, const_3), const_2)) | perimeter of an equilateral and isosceles is 45 and 40 respectively . at least one of the sides of isosceles is equal to the equilateral . what ' s the base of isosceles triangle ? | 10 units as other 2 sides will be 15,15 units . answer : d | a = 45 / 3
b = a * 2
c = 40 - b
|
a ) 65 , b ) 69 , c ) 74 , d ) 75 , e ) none | c | divide(add(add(add(add(76, 60), 82), 67), 85), add(const_1, const_4)) | kamal obtained 76 , 60 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 76 + 60 + 82 + 67 + 85 / 5 ) = ( 375 / 5 ) = 74 . answer c" | a = 76 + 60
b = a + 82
c = b + 67
d = c + 85
e = 1 + 4
f = d / e
|
a ) 320 m , b ) 188 m , c ) 120 m , d ) 178 m , e ) 189 m | a | divide(12, subtract(divide(12, 10), 16)) | a train covers a distance of 12 km in 10 min . if it takes 16 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 16 = 320 m . answer : a" | a = 12 / 10
b = a - 16
c = 12 / b
|
a ) 2 : 9 , b ) 2 : 1 , c ) 2 : 7 , d ) 2 : 5 , e ) 2 : 2 | b | divide(subtract(multiply(4, 24), multiply(5, 16)), subtract(multiply(5, 12), multiply(4, 13))) | if 12 men and 16 boys can do a piece of work in 5 days and 13 men together will 24 boys can do it in 4 days . compare the daily work done by a man with that of a boy . | "12 m + 16 b - - - - - 5 days 13 m + 24 b - - - - - - - 4 days 60 m + 80 b = 52 m + 96 b 8 m = 16 b = > 1 m = 2 b m : b = 2 : 1 answer : b" | a = 4 * 24
b = 5 * 16
c = a - b
d = 5 * 12
e = 4 * 13
f = d - e
g = c / f
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | subtract(431, subtract(431, 431)) | what least value should be replaced by * in 223 * 431 so the number become divisible by 9 | "explanation : trick : number is divisible by 9 , if sum of all digits is divisible by 9 , so ( 2 + 2 + 3 + * + 4 + 3 + 1 ) = 15 + * should be divisible by 9 , 15 + 3 will be divisible by 9 , so that least number is 3 . answer : option a" | a = 431 - 431
b = 431 - a
|
a ) 18 % , b ) 9 % , c ) 22 % , d ) 24 % , e ) 21 | b | multiply(divide(subtract(1340, 1210), 1340), const_100) | the cost price of a radio is rs . 1340 and it was sold for rs . 1210 , find the loss % ? | "1340 - - - - 130 100 - - - - ? = > 9 % answer : b" | a = 1340 - 1210
b = a / 1340
c = b * 100
|
a ) 17 / 65 , b ) 17 / 25 , c ) 17 / 28 , d ) 17 / 65 , e ) 17 / 39 | b | divide(multiply(subtract(multiply(const_6, const_3), const_1), const_2), 50) | the probability that a number selected at random from the first 50 natural numbers is a composite number is ? | "the number of exhaustive events = ⁵ ⁰ c ₁ = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 / 50 = 17 / 25 . answer : b" | a = 6 * 3
b = a - 1
c = b * 2
d = c / 50
|
a ) 7 : 13 , b ) 38 : 7 , c ) 3 : 13 , d ) 2 : 3 , e ) 38 : 13 | e | divide(add(power(6, 2), 2), add(add(4, 7), 2)) | if x : y = 4 : 7 , find the value of ( 6 x + 2 y ) : ( 5 x – y ) | explanation : given : x / y = 4 / 7 ( 6 x + 2 y ) : ( 5 x – y ) = ( 6 * 4 + 2 * 7 ) : ( 5 * 4 – 7 ) 38 : 13 answer : e | a = 6 ** 2
b = a + 2
c = 4 + 7
d = c + 2
e = b / d
|
a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) none of these | a | multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100) | ritesh and co . generated revenue of rs . 2,500 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ? | "explanation : given , ritesh and co . generated revenue of rs . 2,500 in 2006 and that this was 12.5 % of the gross revenue . hence , if 2500 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) × 2500 . = > 20,000 . hence , the total revenue by end of 2007 is rs . 10,000 . in 2006 , revenue grew by rs . 2500 . this is a growth of : = > ( 2500 / 20000 ) × 100 . = > 12.5 % . answer : a" | a = 1 + 4
b = 2 * a
c = b * 100
d = 2 * c
e = 1 + 4
f = e * 100
g = d + f
h = 1 + 4
i = 2 * h
j = i * 100
k = 2 + 4
l = k * 100
m = j + l
n = 12 / 5
o = m / n
p = g / o
q = p * 100
|
a ) 25 , b ) 34 , c ) 50 , d ) 20 , e ) 100 | d | divide(multiply(6, 200), divide(200, const_10)) | according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 15 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 15 oz can = 300 oz / 15 oz = 20 answer d , 20 cans" | a = 6 * 200
b = 200 / 10
c = a / b
|
a ) 6 % , b ) 7 % , c ) 9 % , d ) 8 % , e ) 5 % | d | divide(96, subtract(power(add(const_1, divide(15,000, const_100)), 2), add(const_1, multiply(2, divide(15,000, const_100))))) | the difference between compound interest and simple interest on an amount of $ 15,000 for 2 years is $ 96 . what is the rate of interest per annum ? | "[ 15000 x ( 1 + r / 100 ) ^ 2 - 15000 ] - [ ( 15000 x r x 2 ) / 100 ] = 96 = = > 15000 [ ( 1 + r / 100 ) ^ 2 - 1 - 2 r / 100 ] = 96 = = > 15000 [ ( ( 100 + r ) ^ 2 - 10000 - ( 200 x r ) ) / 10000 ] = 96 = = > r ^ 2 = ( 96 x 2 ) / 3 = 64 = = > r = 8 . rate = 8 % answer d ) 8 %" | a = 15 / 0
b = 1 + a
c = b ** 2
d = 15 / 0
e = 2 * d
f = 1 + e
g = c - f
h = 96 / g
|
a ) 37 , b ) 35 , c ) 25 , d ) 20 , e ) 42 | a | divide(add(14, 60), const_2) | if x + y = 14 , x - y = 60 , for integers of x and y , x = ? | "x + y = 14 x - y = 60 2 x = 74 x = 37 answer is a" | a = 14 + 60
b = a / 2
|
a ) 20 m , b ) 16 m , c ) 11 m , d ) 10 m , e ) 15 m | a | multiply(divide(subtract(25, 20), 25), 100) | if in a race of 100 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | "explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 100 * 5 ) / 25 = 20 m hence , a beats b by 20 m . answer a" | a = 25 - 20
b = a / 25
c = b * 100
|
a ) 50 kmph , b ) 60 kmph , c ) 75 kmph , d ) 85 kmph , e ) 90 kmph | b | divide(add(90, 30), const_2) | the speed of a car is 90 km in the first hour and 30 km in the second hour . what is the average speed of the car ? | "explanation : s = ( 90 + 30 ) / 2 = 60 kmph b )" | a = 90 + 30
b = a / 2
|
a ) $ 10.50 , b ) $ 12.50 , c ) $ 11.50 , d ) $ 16.50 , e ) $ 9.50 | b | divide(subtract(580, multiply(22, divide(subtract(multiply(580, const_2), 800), subtract(multiply(22, const_2), 8)))), 16) | suzie ’ s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 580 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 800 . how much more do pairs of boots cost than pairs of shoes at suzie ’ s discount footwear ? | "let x be pair of shoes and y be pair of boots . 22 x + 16 y = 580 . . . eq 1 8 x + 32 y = 800 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 1160 8 x = 800 . 36 x = 360 = > x = 10 . sub x in eq 2 . . . . we get 80 + 32 y = 800 . . . then we get 32 y = 720 then y = 22.50 differenece between x and y is 12.50 answer : b" | a = 580 * 2
b = a - 800
c = 22 * 2
d = c - 8
e = b / d
f = 22 * e
g = 580 - f
h = g / 16
|
a ) 55.55 km , b ) 44.44 km , c ) 33.33 km , d ) 22.22 km , e ) 11.11 km | c | multiply(add(65, 15), divide(25, const_60)) | the speed of a boat in still water in 65 km / hr and the rate of current is 15 km / hr . the distance travelled downstream in 25 minutes is : | "explanation : speed downstream = ( 65 + 15 ) = 80 kmph time = 25 minutes = 25 / 60 hour = 5 / 12 hour distance travelled = time × speed = ( 2 / 5 ) × 80 = 33.33 km answer : option c" | a = 65 + 15
b = 25 / const_60
c = a * b
|
a ) 2 √ 2 , b ) 2 √ 3 , c ) 3 √ 2 , d ) 3 √ 3 , e ) 3 | e | divide(add(sqrt(27), sqrt(243)), sqrt(48)) | ( √ 27 + √ 243 ) / √ 48 = ? | "( √ 27 + √ 243 ) / √ 48 = ( 3 √ 3 + 9 √ 3 ) / 4 √ 3 = 12 √ 3 / 4 √ 3 = 3 . hence , the correct answer is e ." | a = math.sqrt(27)
b = math.sqrt(243)
c = a + b
d = math.sqrt(48)
e = c / d
|
a ) 2887 , b ) 444 , c ) 877 , d ) 278 , e ) 178 | b | add(multiply(divide(60, subtract(21, 16)), 16), multiply(divide(60, subtract(21, 16)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 16 + 21 = 37 t = 12 d = 37 * 12 = 444 answer : b" | a = 21 - 16
b = 60 / a
c = b * 16
d = 21 - 16
e = 60 / d
f = e * 21
g = c + f
|
a ) 35 % , b ) 36 % , c ) 37 % , d ) 38 % , e ) 40 % | d | subtract(50, divide(50, 8)) | you hold some gold in a vault as an investment . over the past year the price of gold increases by 50 % . in order to keep your gold in the vault , you must pay 8 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | "( 100 % + 50 % ) * ( 100 % - 8 % ) = 150 * 0.92 = 138 % an increase of 38 % your gold holdings have increased in value by 38 % . the answer is d" | a = 50 / 8
b = 50 - a
|
a ) 16.67 % , b ) 17.14 % , c ) 18.3 % , d ) 19.75 % , e ) 21.23 % | b | multiply(divide(subtract(add(18, 3), add(multiply(divide(subtract(const_100, 20), const_100), 18), 3)), add(18, 3)), const_100) | 18 litres of mixture contains 20 % alcohol and the rest water . if 3 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | "alcohol in the 18 litres of mix . = 20 % of 18 litres = ( 20 * 18 / 100 ) = 3.6 litres water in it = 18 - 3.6 = 14.4 litres new quantity of mix . = 18 + 3 = 21 litres quantity of alcohol in it = 3.6 litres percentage of alcohol in new mix . = 3.6 * 100 / 21 = 17.14 % answer is b" | a = 18 + 3
b = 100 - 20
c = b / 100
d = c * 18
e = d + 3
f = a - e
g = 18 + 3
h = f / g
i = h * 100
|
['a ) 4 / 5', 'b ) 5 / 4', 'c ) 3 / 2', 'd ) 5 / 7', 'e ) 2 / 3'] | e | divide(sqrt(const_4), sqrt(9)) | two isosceles triangles have equal vertical angles and their areas are in the ratio 4 : 9 . find the ratio of their corresponding heights . | we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 4 / 9 - - > height / height = 2 / 3 . answer : e . | a = math.sqrt(4)
b = math.sqrt(9)
c = a / b
|
a ) 6 , b ) 21 , c ) 27 , d ) 189 , e ) 192 | d | divide(multiply(divide(3, subtract(divide(8, 9), divide(7, 8))), 7), 8) | in a certain company , the ratio of male to female employees is 7 : 8 . if 3 more men were hired , this ratio would increase to 8 : 9 . how many male employees are there in the company ? | another approach is to use two variables . let m = present number of males let f = present number of females the ratio of male to female employees is 7 : 8 so , m / f = 7 / 8 cross multiply to get 7 f = 8 m if 3 more men were hired , this ratio would increase to 8 : 9 so , ( m + 3 ) / f = 8 / 9 cross multiply to get 9 ( m + 3 ) = 8 f expand : 9 m + 27 = 8 f we now have a system of two equations and two variables : 7 f = 8 m 9 m + 27 = 8 f solve to get : m = 189 and f = 216 answer : d | a = 8 / 9
b = 7 / 8
c = a - b
d = 3 / c
e = d * 7
f = e / 8
|
a ) 18 , b ) 15 , c ) 33 , d ) 39 , e ) 50 | e | add(multiply(5, 5), multiply(5, 5)) | a guy was asked to specify his age in years . he said , “ take my age 5 years hence , multiply it by 5 and subtract 5 times of my age 5 years ago and you will know my age . ” what was the age of that guy ? | current age of the guy = a years . then , 5 ( a + 5 ) – 5 ( a – 5 ) = a ( 5 a + 25 ) – ( 5 a – 25 ) = a a = 50 e | a = 5 * 5
b = 5 * 5
c = a + b
|
a ) 5 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(subtract(17, subtract(10, 4)), 6) | of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 17 boys and 43 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ? | "venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you will use logic : there are 6 happy boys . there are 4 sad girls but total 10 sad children . so rest 6 sad children must be sad boys . we have 6 happy boys and 6 sad boys . total we have 17 boys . so 17 - 6 - 6 = 5 boys must be neither happy nor sad . answer ( a )" | a = 10 - 4
b = 17 - a
c = b - 6
|
a ) 16 days , b ) 12 days , c ) 9 days , d ) 6 days , e ) 18 days | c | inverse(add(divide(const_1, 27), divide(const_1, divide(27, const_2)))) | ram , who is half as efficient as krish , will take 27 days to complete a task if he worked alone . if ram and krish worked together , how long will they take to complete the task ? | number of days taken by ram to complete task = 27 since ram is half as efficient as krish , amount of work done by krish in 1 day = amount of work done by ram in 2 days if total work done by ram in 27 days is 27 w amount of work done by ram in 1 day = w amount of work done by krish in 1 day = 2 w total amount of work done by krish and ram in a day = 3 w total amount of time needed by krish and ram to complete task = 27 w / 3 w = 9 days answer c | a = 1 / 27
b = 27 / 2
c = 1 / b
d = a + c
e = 1/(d)
|
a ) 1058.24 , b ) 2006.24 , c ) 1006.13 , d ) 1015.24 , e ) 1014.24 | c | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(4, const_3)))), const_100) | what is the compound interest on rs : 60,000 for 4 months at the rate of 5 % per annum | "it is monthly compound rate = 5 / 12 % per month 60000 * ( 1 + 5 / 1200 ) ^ 4 - 60000 = 1006.13 answer : c" | a = 3 * 100
b = a * 100
c = 4 * 3
d = 4 / c
e = 5 * d
f = b * e
g = f / 100
|
a ) 1 / 8 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 1 / 4 | e | multiply(6, add(divide(const_1, 40), divide(const_1, 60))) | two persons a and b can complete a piece of work in 40 days and 60 days respectively . if they work together , what part of the work will be completed in 6 days ? | "a ' s one day ' s work = 1 / 40 b ' s one day ' s work = 1 / 60 ( a + b ) ' s one day ' s work = 1 / 40 + 1 / 60 = 1 / 24 the part of the work completed in 6 days = 6 ( 1 / 24 ) = 1 / 4 . answer : e" | a = 1 / 40
b = 1 / 60
c = a + b
d = 6 * c
|
a ) 130 , b ) 125 , c ) 145 , d ) 135 , e ) 144 | b | divide(multiply(add(95, divide(multiply(95, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 95 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 95 sp = 47.50 * ( 125 / 100 ) = 118.125 mp * ( 95 / 100 ) = 118.125 mp = 125 answer : b" | a = 95 * 25
b = a / 100
c = 95 + b
d = c * 100
e = 100 - 5
f = d / e
|
a ) 3 , b ) 7 , c ) 21 , d ) 189 , e ) 267 | d | multiply(multiply(multiply(subtract(add(add(subtract(9, 1), add(2, 1)), add(2, 1)), const_10), subtract(add(multiply(2, add(subtract(9, 1), add(2, 1))), 1), const_10)), add(2, 1)), subtract(add(multiply(2, add(add(add(subtract(9, 1), add(2, 1)), add(2, 1)), add(2, 1))), 1), multiply(2, const_10))) | a “ sophie germain ” prime is any positive prime number p for which 2 p + 1 is also prime . the product of all the possible units digits of sophie germain primes greater than 9 is | "in that case , the sophie prime numbers greater than 9 are 11,23 , 47,59 , . . which yields units digit as 1 , 3,7 and 9 product would be 1 x 3 x 7 x 9 = 189 answer should be d" | a = 9 - 1
b = 2 + 1
c = a + b
d = 2 + 1
e = c + d
f = e - 10
g = 9 - 1
h = 2 + 1
i = g + h
j = 2 * i
k = j + 1
l = k - 10
m = f * l
n = 2 + 1
o = m * n
p = 9 - 1
q = 2 + 1
r = p + q
s = 2 + 1
t = r + s
u = 2 + 1
v = t + u
w = 2 * v
x = w + 1
y = 2 * 10
z = x - y
A = o * z
|
a ) 18 km , b ) 10 km , c ) 12 km , d ) 24 km , e ) 25 km | a | multiply(3, 3) | a man performs 1 / 2 of the total journey by rail , 1 / 3 by bus and the remaining 3 km on foot . his total journey is | "explanation : let the journey be x km then , 1 x / 2 + 1 x / 3 + 3 = x 5 x + 18 = 6 x x = 18 km answer : option a" | a = 3 * 3
|
a ) 110 , b ) 135 , c ) 150 , d ) 165 , e ) 235 | d | add(add(add(add(27, const_2), add(add(27, const_2), const_2)), add(add(add(27, const_2), const_2), const_2)), 37) | the sum of all consecutive odd integers from − 27 to 37 , inclusive , is | "the sum of the odd numbers from - 27 to + 27 is 0 . let ' s add the remaining numbers . 29 + 31 + 33 + 35 + 37 = 5 ( 33 ) = 165 the answer is d ." | a = 27 + 2
b = 27 + 2
c = b + 2
d = a + c
e = 27 + 2
f = e + 2
g = f + 2
h = d + g
i = h + 37
|
a ) 2 : 5 , b ) 1 : 4 , c ) 5 : 3 , d ) 6 : 11 , e ) 2 : 3 | c | divide(divide(const_1, const_4), divide(20, const_100)) | if 20 % of a number is equal to one - third of another number , what is the ratio of first number to the second number ? | "let 20 % of a = 1 / 3 b then 20 a / 100 = 1 b / 3 a / 5 = b / 3 a / b = 5 / 3 a : b = 5 : 3 answer is c" | a = 1 / 4
b = 20 / 100
c = a / b
|
a ) 33 , b ) 75 , c ) 48 , d ) 99 , e ) 21 | b | divide(add(multiply(60, add(const_1, divide(50, const_100))), 60), const_2) | a person travels from p to q a speed of 60 km / hr and returns by increasing his speed by 50 % . what is his average speed for both the trips ? | speed on return trip = 150 % of 60 = 90 km / hr . average speed of trip = 60 + 90 / 2 = 150 / 2 = 75 km / hr answer : b | a = 50 / 100
b = 1 + a
c = 60 * b
d = c + 60
e = d / 2
|
a ) 54 , b ) 33 , c ) 44 , d ) 22 , e ) 26 | c | subtract(divide(multiply(1.1, add(1.1, const_1)), const_2), divide(multiply(subtract(1.1, const_1), 1.1), const_2)) | ( 1.1 + 1.1 + 1.1 + 1.1 ) x 1.1 x 1.1 = ? = ? x 0.121 | "explanation : ? = ( 1.1 + 1.1 + 1.1 + 1.1 ) x 1.1 x 1.1 / 0.121 = ( 4.4 x 1.1 x 1.1 ) / 0.121 = 44 answer : option c" | a = 1 + 1
b = 1 * 1
c = b / 2
d = 1 - 1
e = d * 1
f = e / 2
g = c - f
|
a ) 22 , b ) 14 , c ) 88 , d ) 12 , e ) 66 | b | subtract(multiply(40, divide(85, const_100)), multiply(divide(4, 5), 25)) | how much is 85 % of 40 is greater than 4 / 5 of 25 ? | "( 85 / 100 ) * 40 – ( 4 / 5 ) * 25 = 14 answer : b" | a = 85 / 100
b = 40 * a
c = 4 / 5
d = c * 25
e = b - d
|
a ) 1680579288 , b ) 1223441288 , c ) 2142579288 , d ) 2142339288 , e ) none of these | a | multiply(23341379, power(add(const_4, const_1), const_4)) | ( 23341379 x 72 ) = ? | "explanation : 23341379 x 72 = 23341379 ( 70 + 2 ) = ( 23341379 x 70 ) + ( 23341379 x 2 ) = 1633896530 + 46682758 = 1680579288 . answer : option a" | a = 4 + 1
b = a ** 4
c = 23341379 * b
|
a ) $ 338.50 , b ) $ 341.40 , c ) $ 344.30 , d ) $ 347.20 , e ) $ 350.10 | d | add(add(multiply(multiply(add(divide(multiply(120, 65), const_100), 65), 2), 0.7), multiply(multiply(3, 65), 0.7)), multiply(2.1, add(3, 2))) | in a fuel station the service costs $ 2.10 per vehicle and every liter of fuel costs $ 0.70 . assuming that you fill up 3 mini - vans and 2 trucks , what will be the total cost , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty ? | the service cost of 3 vans and 2 trucks is 5 * 2.10 $ 10.50 the fuel in 3 vans is 3 * 65 = 195 liters the fuel in 2 trucks is 2 * 65 * 2.2 = 286 liters the total fuel ( vans + trucks ) = 481 liters the total fuel cost is 481 * 0.7 = $ 336.70 the total cost is $ 336.70 + $ 10.50 = $ 347.20 the answer is d . | a = 120 * 65
b = a / 100
c = b + 65
d = c * 2
e = d * 0
f = 3 * 65
g = f * 0
h = e + g
i = 3 + 2
j = 2 * 1
k = h + j
|
a ) 11 . , b ) 12 . , c ) 13 . , d ) 18 . , e ) 14.5 | d | add(divide(multiply(10, const_100), add(const_100, 25)), divide(multiply(15, const_100), add(const_100, 50))) | following an increase in prices , the price of a candy box was 10 pounds and the price of a can of soda was 15 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | price of candy before price increase = 10 / 1.25 = 8 price of soda before price increase = 15 / 1.5 = 10 total price = 8 + 10 = 18 d is the answer | a = 10 * 100
b = 100 + 25
c = a / b
d = 15 * 100
e = 100 + 50
f = d / e
g = c + f
|
a ) 512 , b ) 768 , c ) 4096 , d ) 2048 , e ) 1024 | e | subtract(power(2, add(9, const_1)), const_1) | the population of a bacteria colony doubles every day . if it was started 9 days ago with 2 bacteria and each bacteria lives for 11 days , how large is the colony today ? | "9 days ago - 2 8 days ago - 4 7 days ago - 8 6 days ago - 16 5 days ago - 32 4 days ago - 64 3 days ago - 128 2 days ago - 256 yesterday - 512 today - 1024 ans : e" | a = 9 + 1
b = 2 ** a
c = b - 1
|
a ) 120 , b ) 190 , c ) 224 , d ) 298 , e ) 256 | c | multiply(subtract(power(2, 4), 2), multiply(4, 4)) | in how many ways can an answer key for a quiz be written if the quiz contains 4 true - false questions followed by 2 multiples - choice questions with 4 answer choices each , if the correct answers to all true - false questions can not be the same ? | "there are 2 ^ 4 = 16 possibilities for the true - false answers . however we need to remove two cases for tttt and ffff . there are 4 * 4 = 16 possibilities for the multiple choice questions . the total number of possibilities is 14 * 16 = 224 . the answer is c ." | a = 2 ** 4
b = a - 2
c = 4 * 4
d = b * c
|
a ) 3.5 , b ) 2.7 , c ) 2.9 , d ) 2.3 , e ) 2.1 | a | divide(135, multiply(140, const_0_2778)) | in what time will a train 135 m long cross an electric pole , it its speed be 140 km / hr ? | "speed = 140 * 5 / 18 = 38.8 m / sec time taken = 135 / 38.8 = 3.5 sec . answer : a" | a = 140 * const_0_2778
b = 135 / a
|
a ) s . 1080 , b ) s . 1140 , c ) s . 1090 , d ) s . 1202 , e ) s . 1092 | a | divide(multiply(subtract(const_100, 10), 1200), const_100) | a man buys a cycle for rs . 1200 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 1200 = rs . 90 x 1200 / 100 = rs . 1080 answer : option a" | a = 100 - 10
b = a * 1200
c = b / 100
|
a ) 43 , b ) 44 , c ) 45 , d ) 47 , e ) 48 | c | subtract(add(add(multiply(30, 14), 15), 30), multiply(30, 14)) | the average age of 30 students in a class is 14 years . if the age of teacher is also included , the average becomes 15 years , find the age of the teacher . | explanation : if teacher ' s age is 14 years , there is no change in the average . but teacher has contributed 1 year to all the students along with maintaining his age at 15 . age of teacher = average age of all + total increase in age = 15 + ( 1 x 30 ) = 45 years answer : c | a = 30 * 14
b = a + 15
c = b + 30
d = 30 * 14
e = c - d
|
a ) 150 miles , b ) 160 miles , c ) 140 miles , d ) 170 miles , e ) 200 miles | b | multiply(40, divide(240, 60)) | john and lewis leave city a for city b simultaneously at 6 a . m in the morning driving in two cars at speeds of 40 mph and 60 mph respectively . as soon as lewis reaches city b , he returns back to city a along the same route and meets john on the way back . if the distance between the two cities is 240 miles , how far from city a did john and lewis meet ? | time taken by lewis to reach city b = 240 / 60 = 4 hours in 4 hours , john travels 40 * 4 = 160 miles so distance at which they meet should be greater than 160 miles . only b satisfies . answer is b . | a = 240 / 60
b = 40 * a
|
a ) s . 8,000 , b ) s . 9,000 , c ) s . 20,000 , d ) s . 10,000 , e ) s . 50,000 | a | divide(divide(subtract(subtract(1000, multiply(4000, divide(2.5, const_100))), 700), subtract(divide(5, const_100), divide(2.5, const_100))), 1000) | a salesman â € ™ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4000 . if his remuneration as per new scheme was rs . 700 more than that by the previous schema , his sales were worth ? | [ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 700 x = 8000 answer a | a = 2 / 5
b = 4000 * a
c = 1000 - b
d = c - 700
e = 5 / 100
f = 2 / 5
g = e - f
h = d / g
i = h / 1000
|
a ) 75 kg , b ) 65 kg , c ) 55 kg , d ) 89 kg , e ) 25 kg | d | add(65, multiply(8, 3)) | the average weight of 8 persons increases by 3 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | total weight increased = ( 8 x 3 ) kg = 24 kg . weight of new person = ( 65 + 24 ) kg = 89 kg . answer : d | a = 8 * 3
b = 65 + a
|
a ) 65 , b ) 55 , c ) 45 , d ) 35 , e ) 25 | e | subtract(55, subtract(100, add(subtract(100, 55), add(subtract(100, 85), subtract(100, 90))))) | in a village of 100 households , 85 have at least one dvd player , 90 have at least one cell phone , and 55 have at least one mp 3 player . if x and y are respectively the greatest and lowest possible number of households that have all 3 of these devices , x – y is : | am i missing something here ? ? ? it seems straightforward . . . . . . the obvious maximum that have all 3 is 55 , because you are limited by the smallest number . the minimum is simply the sum of the max of each people who dont have the product , so : 100 - 90 = 10 do n ' t have cell 100 - 85 = 15 do n ' t have dvd and 100 - 55 = 45 do n ' t have mp 3 so a total of 10 + 15 + 45 = 70 combined who might not have some combination of the 3 products . so subtract that from 100 , to give you the minimum of the people who could have all 3 and you get 100 - 70 = 30 . 55 - 30 = 25 e | a = 100 - 55
b = 100 - 85
c = 100 - 90
d = b + c
e = a + d
f = 100 - e
g = 55 - f
|
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